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无代写-ECE 272A

时间：2021-04-28

Stochastic Processes in Dynamic Systems

ECE 272A

Midterm Exam-Solution

Instructor: Behrouz Touri

1. For each of the following statements, determine if the statement is true or false. If

you believe the statement is wrong, provide a counterexample, otherwise, prove the

statement.

a. If a linear system x˙ = A(t)x has a solution for some x(0) = x0 ∈ Rn, the solution is

always unique.

b. A dynamics x˙ = f(x) cannot have multiple marginally stable equilibrium.

Solution:

a. False.

Consider the one-dimensional linear system defined by x˙ = A(t)x, where

A(t) =

{

2

t ∀ t > 0

0 t = 0

and let x(0) = 0. For this linear system you can verify that x(t) = 0, for all t, is a

solution. Also, note that x(t) = t2 for all t ≥ 0 is another solution.

b. False.

The dynamics x˙ = f(x) where f(x) = 0 for all x ∈ R has multiple marginally stable

equilibrium (namely all the points on the real line).

2. A model for the unicycle’s movement on the plain whose linear velocity is v and angular

velocity is ω is given by

p˙ =

(

v cos(θ)

v sin(θ)

)

, θ˙ = ω, (1)

where p(t) ∈ R2 is the the Cartesian coordinates of unicycle and θ its orientation. Let

the input to this system be u =

(

v

ω

)

∈ R2.

(a) Let

x =

x1x2

x3

=

p1 cos(θ) + (p2 − 1) sin(θ)−p1 sin(θ) + (p2 − 1) cos(θ)

θ

.

Write down the state-space formulation of this system using state variable x and

input variable u, i.e., find out f(x, u) such that x˙ = f(x, u).

1

Solution: By calculating x˙ and plugging in p˙ and θ˙, we have

x˙ =

p˙1 cos(θ)− p1θ˙ sin(θ) + p˙2 sin(θ) + (p2 − 1)θ˙ cos(θ)−p˙1 sin(θ) + p1θ˙ cos(θ) + p˙2 cos(θ)− (p2 − 1)θ˙ sin(θ)

θ˙

=

v − p1ω sin(θ) + (p2 − 1)ω cos(θ)p1ω cos(θ)− (p2 − 1)ω sin(θ)

ω

=

v + x2ω−x1ω

ω

.

(b) Show that the origin (x∗ = (0, 0, 0)T and u∗ = (0, 0)T ) is an equilibirum for the

dynamics in part (a). Find the linearization x˙ = Ax+Bu around this equilibrium.

Solution: Since

f(x∗, u∗) =

v∗ + x∗2ω∗−x∗1ω∗

ω∗

=

00

0

,

the origin is an equilibrium. We have

∇xf(x, u) =

0 ω 0−ω 0 0

0 0 0

,∇uf(x, u) =

1 x20 −x1

0 1

.

Therefore, the linearization around the equilibrium is

x˙ = ∇xf(x∗, u∗)x+∇uf(x∗, u∗)u =

0 0 00 0 0

0 0 0

x+

1 00 0

0 1

u.

(c) Show that ω(t) = 1 = v(t) and p(t) =

(

sin(t)

1− cos(t)

)

and θ(t) = t is a solution

for the original dynamics (1). Find the linearization of the dynamics in Part (a)

around this trajectory.

Solution: We have

x =

p1 cos(θ) + (p2 − 1) sin(θ)−p1 sin(θ) + (p2 − 1) cos(θ)

θ

=

0−1

t

, u = (v

ω

)

=

(

1

1

)

.

Therefore,

x˙ =

d

dt

0−1

t

=

00

1

.

2

This satisfies

x˙ =

v + x2ω−x1ω

ω

=

1 + (−1)1−0 · 1

1

=

00

1

.

Hence this is a solution. The linearization around the trajectory is

x˙ = ∇xf(x∗(t), u∗(t))x+∇uf(x∗(t), u∗(t))u =

0 1 0−1 0 0

0 0 0

x+

1 −10 0

0 1

u.

3. We say that a linear time-varying system x˙ = A(t)x(t) is preserving direction u 6= 0 ∈ Rn

if A(t)u = λ(t)u, where n×n matrix A(t) and λ(t) ∈ R are continuous functions of time

t ≥ 0.

(a) Find the solution of the system starting at the initial condition x(0) = αu for some

α ∈ R.

Solution: Suppose x(t) = g(t)u, where g(t) ∈ R, g(0) = α. Since x˙ = A(t)x(t), we

have

g˙(t)u = A(t)g(t)u = g(t)A(t)u = g(t)λ(t)u.

Since u 6= 0, g˙(t) = g(t)λ(t). Therefore,

g(t) = g(0) exp

( ∫ t

0

λ(t)dt

)

= α exp

( ∫ t

0

λ(t)dt

)

.

Now, since x(t) = αu exp

( ∫ t

0 λ(t)dt

)

satisfies x˙ = A(t)x(t) and the initial condition

and A(t) is a continuous function, it is the solution to the ODE.

(b) Show that for all α 6= 0 in part (a), we have limt→∞ ‖x(t)‖ = 0, if and only if∫∞

0 λ(t)dt = −∞.

Solution: We have

‖x(t)‖ = ‖αu exp ( ∫ t

0

λ(t)dt

)‖ = ‖αu‖ exp ( ∫ t

0

λ(t)dt

)

,

and ‖αu‖ > 0 since α 6= 0 and u 6= 0. Therefore, if ∫∞0 λ(t)dt = −∞, then

lim

t→∞ ‖x(t)‖ = ‖αu‖ limt→∞ exp

( ∫ t

0

λ(t)dt

)

= 0.

Also, if limt→∞ ‖x(t)‖ = 0, then

lim

t→∞ exp

( ∫ t

0

λ(t)dt

)

=

1

‖αu‖ limt→∞ ‖x(t)‖ = 0,

which implies

∫∞

0 λ(t)dt = −∞.

3

4. Consider the dynamics:

x˙ =

(−1 1

−2 −3

)

x+

(

0

1

)

u(t).

a. Find the solution for the unforced system with the initial condition x(0) = (−1, 1)T .

Solution: By eigenvalue decomposition, we have(−1 1

−2 −3

)

=

(−1 + i −1− i

2 2

)(−2− i 0

0 −2 + i

)(−i

2

1−i

4

i

2

1+i

4

)

.

Therefore, the solution is

x(t) = eAtx(0)

=

(−1 + i −1− i

2 2

)(

e(−2−i)t 0

0 e(−2+i)t

)(−i

2

1−i

4

i

2

1+i

4

)(−1

1

)

=

( −12e(−2−i)t − 12e(−2+i)t

1+i

2 e

(−2−i)t + 1−i2 e

(−2+i)t

)

=

( −e−2t cos t

e−2t cos t+ e−2t sin t

)

.

Solution: By Laplace transform,

X(t) = (sI −A)−1x(0)

=

1

s2 + 4s+ 5

(

s+ 3 1

−2 s+ 1

)

x(0)

=

1

(s+ 2)2 + 1

(−s− 2

s+ 3

)

.

By Inverse Laplace transform,

x(t) =

( −e−2t cos t

e−2t cos t+ e−2t sin t

)

.

b. Determine whether the origin for the system is marginally stable, asymptotically

stable, or unstable. Find the stability properties of the (equilibrium) origin for the

unforced system.

Solution: Let λ1 = −2 − i, λ2 = −2 + i. Since Re[λi] < 0 for i = 1, 2, the origin is

asymptotically (and exponentially) stable.

c. Show that if the input u(t) is an integrable and uniformly bounded function, i.e.,

|u(t)| ≤ β for some β > 0 and all t ≥ 0, then the trajectory is also uniformly

bounded, i.e., ‖x(t)‖ ≤ βˆ for some βˆ > 0 and all t ≥ 0.

Solution: The solution is

x(t) = eAtx(0) +

∫ t

0

eA(t−τ)

(

0

1

)

u(τ)dτ.

4

Therefore,

‖x(t)‖ ≤ ‖eAtx(0)‖+

∥∥∥∥∫ t

0

eA(t−τ)

(

0

1

)

u(τ)dτ

∥∥∥∥

≤ ‖eAtx(0)‖+

∫ t

0

∥∥∥∥eA(t−τ)(01

)

u(τ)

∥∥∥∥dτ

≤ ‖eAtx(0)‖+

∫ t

0

‖eA(t−τ)‖

∥∥∥∥(01

)∥∥∥∥|u(τ)|dτ

≤ ‖eAtx(0)‖+

∫ t

0

‖eA(t−τ)‖ · 1 · βdτ

= ‖eAtx(0)‖+ β

∫ t

0

‖eλ1(t−τ)P1 + eλ2(t−τ)P2‖dτ

≤ ‖eAtx(0)‖+ β‖P1‖

∫ t

0

e−2(t−τ)dτ + β‖P2‖

∫ t

0

e−2(t−τ)dτ

= ‖eAtx(0)‖+ β(‖P1‖+ ‖P2‖)1− e

−2t

2

,

where A = λ1P1 + λ2P2 for some P1, P2. Since the first term converges to 0 and the

second is bounded for all t ≥ 0, ‖x(t)‖ is uniformly bounded.

5

学霸联盟

ECE 272A

Midterm Exam-Solution

Instructor: Behrouz Touri

1. For each of the following statements, determine if the statement is true or false. If

you believe the statement is wrong, provide a counterexample, otherwise, prove the

statement.

a. If a linear system x˙ = A(t)x has a solution for some x(0) = x0 ∈ Rn, the solution is

always unique.

b. A dynamics x˙ = f(x) cannot have multiple marginally stable equilibrium.

Solution:

a. False.

Consider the one-dimensional linear system defined by x˙ = A(t)x, where

A(t) =

{

2

t ∀ t > 0

0 t = 0

and let x(0) = 0. For this linear system you can verify that x(t) = 0, for all t, is a

solution. Also, note that x(t) = t2 for all t ≥ 0 is another solution.

b. False.

The dynamics x˙ = f(x) where f(x) = 0 for all x ∈ R has multiple marginally stable

equilibrium (namely all the points on the real line).

2. A model for the unicycle’s movement on the plain whose linear velocity is v and angular

velocity is ω is given by

p˙ =

(

v cos(θ)

v sin(θ)

)

, θ˙ = ω, (1)

where p(t) ∈ R2 is the the Cartesian coordinates of unicycle and θ its orientation. Let

the input to this system be u =

(

v

ω

)

∈ R2.

(a) Let

x =

x1x2

x3

=

p1 cos(θ) + (p2 − 1) sin(θ)−p1 sin(θ) + (p2 − 1) cos(θ)

θ

.

Write down the state-space formulation of this system using state variable x and

input variable u, i.e., find out f(x, u) such that x˙ = f(x, u).

1

Solution: By calculating x˙ and plugging in p˙ and θ˙, we have

x˙ =

p˙1 cos(θ)− p1θ˙ sin(θ) + p˙2 sin(θ) + (p2 − 1)θ˙ cos(θ)−p˙1 sin(θ) + p1θ˙ cos(θ) + p˙2 cos(θ)− (p2 − 1)θ˙ sin(θ)

θ˙

=

v − p1ω sin(θ) + (p2 − 1)ω cos(θ)p1ω cos(θ)− (p2 − 1)ω sin(θ)

ω

=

v + x2ω−x1ω

ω

.

(b) Show that the origin (x∗ = (0, 0, 0)T and u∗ = (0, 0)T ) is an equilibirum for the

dynamics in part (a). Find the linearization x˙ = Ax+Bu around this equilibrium.

Solution: Since

f(x∗, u∗) =

v∗ + x∗2ω∗−x∗1ω∗

ω∗

=

00

0

,

the origin is an equilibrium. We have

∇xf(x, u) =

0 ω 0−ω 0 0

0 0 0

,∇uf(x, u) =

1 x20 −x1

0 1

.

Therefore, the linearization around the equilibrium is

x˙ = ∇xf(x∗, u∗)x+∇uf(x∗, u∗)u =

0 0 00 0 0

0 0 0

x+

1 00 0

0 1

u.

(c) Show that ω(t) = 1 = v(t) and p(t) =

(

sin(t)

1− cos(t)

)

and θ(t) = t is a solution

for the original dynamics (1). Find the linearization of the dynamics in Part (a)

around this trajectory.

Solution: We have

x =

p1 cos(θ) + (p2 − 1) sin(θ)−p1 sin(θ) + (p2 − 1) cos(θ)

θ

=

0−1

t

, u = (v

ω

)

=

(

1

1

)

.

Therefore,

x˙ =

d

dt

0−1

t

=

00

1

.

2

This satisfies

x˙ =

v + x2ω−x1ω

ω

=

1 + (−1)1−0 · 1

1

=

00

1

.

Hence this is a solution. The linearization around the trajectory is

x˙ = ∇xf(x∗(t), u∗(t))x+∇uf(x∗(t), u∗(t))u =

0 1 0−1 0 0

0 0 0

x+

1 −10 0

0 1

u.

3. We say that a linear time-varying system x˙ = A(t)x(t) is preserving direction u 6= 0 ∈ Rn

if A(t)u = λ(t)u, where n×n matrix A(t) and λ(t) ∈ R are continuous functions of time

t ≥ 0.

(a) Find the solution of the system starting at the initial condition x(0) = αu for some

α ∈ R.

Solution: Suppose x(t) = g(t)u, where g(t) ∈ R, g(0) = α. Since x˙ = A(t)x(t), we

have

g˙(t)u = A(t)g(t)u = g(t)A(t)u = g(t)λ(t)u.

Since u 6= 0, g˙(t) = g(t)λ(t). Therefore,

g(t) = g(0) exp

( ∫ t

0

λ(t)dt

)

= α exp

( ∫ t

0

λ(t)dt

)

.

Now, since x(t) = αu exp

( ∫ t

0 λ(t)dt

)

satisfies x˙ = A(t)x(t) and the initial condition

and A(t) is a continuous function, it is the solution to the ODE.

(b) Show that for all α 6= 0 in part (a), we have limt→∞ ‖x(t)‖ = 0, if and only if∫∞

0 λ(t)dt = −∞.

Solution: We have

‖x(t)‖ = ‖αu exp ( ∫ t

0

λ(t)dt

)‖ = ‖αu‖ exp ( ∫ t

0

λ(t)dt

)

,

and ‖αu‖ > 0 since α 6= 0 and u 6= 0. Therefore, if ∫∞0 λ(t)dt = −∞, then

lim

t→∞ ‖x(t)‖ = ‖αu‖ limt→∞ exp

( ∫ t

0

λ(t)dt

)

= 0.

Also, if limt→∞ ‖x(t)‖ = 0, then

lim

t→∞ exp

( ∫ t

0

λ(t)dt

)

=

1

‖αu‖ limt→∞ ‖x(t)‖ = 0,

which implies

∫∞

0 λ(t)dt = −∞.

3

4. Consider the dynamics:

x˙ =

(−1 1

−2 −3

)

x+

(

0

1

)

u(t).

a. Find the solution for the unforced system with the initial condition x(0) = (−1, 1)T .

Solution: By eigenvalue decomposition, we have(−1 1

−2 −3

)

=

(−1 + i −1− i

2 2

)(−2− i 0

0 −2 + i

)(−i

2

1−i

4

i

2

1+i

4

)

.

Therefore, the solution is

x(t) = eAtx(0)

=

(−1 + i −1− i

2 2

)(

e(−2−i)t 0

0 e(−2+i)t

)(−i

2

1−i

4

i

2

1+i

4

)(−1

1

)

=

( −12e(−2−i)t − 12e(−2+i)t

1+i

2 e

(−2−i)t + 1−i2 e

(−2+i)t

)

=

( −e−2t cos t

e−2t cos t+ e−2t sin t

)

.

Solution: By Laplace transform,

X(t) = (sI −A)−1x(0)

=

1

s2 + 4s+ 5

(

s+ 3 1

−2 s+ 1

)

x(0)

=

1

(s+ 2)2 + 1

(−s− 2

s+ 3

)

.

By Inverse Laplace transform,

x(t) =

( −e−2t cos t

e−2t cos t+ e−2t sin t

)

.

b. Determine whether the origin for the system is marginally stable, asymptotically

stable, or unstable. Find the stability properties of the (equilibrium) origin for the

unforced system.

Solution: Let λ1 = −2 − i, λ2 = −2 + i. Since Re[λi] < 0 for i = 1, 2, the origin is

asymptotically (and exponentially) stable.

c. Show that if the input u(t) is an integrable and uniformly bounded function, i.e.,

|u(t)| ≤ β for some β > 0 and all t ≥ 0, then the trajectory is also uniformly

bounded, i.e., ‖x(t)‖ ≤ βˆ for some βˆ > 0 and all t ≥ 0.

Solution: The solution is

x(t) = eAtx(0) +

∫ t

0

eA(t−τ)

(

0

1

)

u(τ)dτ.

4

Therefore,

‖x(t)‖ ≤ ‖eAtx(0)‖+

∥∥∥∥∫ t

0

eA(t−τ)

(

0

1

)

u(τ)dτ

∥∥∥∥

≤ ‖eAtx(0)‖+

∫ t

0

∥∥∥∥eA(t−τ)(01

)

u(τ)

∥∥∥∥dτ

≤ ‖eAtx(0)‖+

∫ t

0

‖eA(t−τ)‖

∥∥∥∥(01

)∥∥∥∥|u(τ)|dτ

≤ ‖eAtx(0)‖+

∫ t

0

‖eA(t−τ)‖ · 1 · βdτ

= ‖eAtx(0)‖+ β

∫ t

0

‖eλ1(t−τ)P1 + eλ2(t−τ)P2‖dτ

≤ ‖eAtx(0)‖+ β‖P1‖

∫ t

0

e−2(t−τ)dτ + β‖P2‖

∫ t

0

e−2(t−τ)dτ

= ‖eAtx(0)‖+ β(‖P1‖+ ‖P2‖)1− e

−2t

2

,

where A = λ1P1 + λ2P2 for some P1, P2. Since the first term converges to 0 and the

second is bounded for all t ≥ 0, ‖x(t)‖ is uniformly bounded.

5

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